Problem 28 - Differential Equations and Linear Algebra

Question

Assume that \(T\) defines a linear transformation and use the given information to find the matrix of \(T.\) \(T: \mathbb{R}^{4} \rightarrow \mathbb{R}^{2}\) such that \(T(1,0,0,0)=(3,-2)\) \(T(1,1,0,0)=(5,1), T(1,1,1,0)=(-1,0),\) and \(T(1,1,1,1)=(2,2)\).

Step-by-Step Solution

Answer

Given the linear transformation \(T : \mathbb{R}^4 \rightarrow \mathbb{R}^2\), we find the matrix representation \(A\) by using the given transformations and the linearity properties. We obtain the column vectors of the matrix \(A = \begin{pmatrix} 3 & 2 & -6 & 3 \\ -2 & 3 & -1 & 2 \end{pmatrix}\). Therefore, the matrix representation of the transformation T is \(T(v) = \begin{pmatrix} 3 & 2 & -6 & 3 \\ -2 & 3 & -1 & 2 \end{pmatrix}v\).

Step 1: Understand the linear transformation

Since \(T\) is a linear transformation, we know it must satisfy the following properties: 1. \(T(u+v) = T(u) + T(v)\) for any vectors u and v. 2. \(T(cv) = cT(v)\) for any vector v and scalar c. Moreover, we know that a matrix A represents a linear transformation T if \(T(v) = Av\) for all vectors v. Our task is to find the matrix representation A of the given transformation T.

Step 2: Finding the column vectors of matrix A

The given information allows us to write the following equations: 1. \(T(1, 0, 0, 0) = (3, -2) \Rightarrow T(e_1) = (3, -2)\) 2. \(T(1, 1, 0, 0) = (5, 1) \Rightarrow T(e_1 + e_2) = (5, 1)\) 3. \(T(1, 1, 1, 0) = (-1, 0) \Rightarrow T(e_1 + e_2 + e_3) = (-1, 0)\) 4. \(T(1, 1, 1, 1) = (2,2) \Rightarrow T(e_1 + e_2 + e_3 + e_4) = (2,2)\) where \(e_1 = (1,0,0,0), e_2 = (0,1,0,0), e_3 = (0,0,1,0)\) and \(e_4 = (0,0,0,1)\). To find column vectors of matrix A, we will use the linear transformation properties. Since T(e_1) gives us the first column, we have: 1. First column: \(A_1 = (3, -2)\) Now let's find the other columns, using the given information: 2. \(T(e_2) = T((1,1,0,0) - (1,0,0,0)) = T(1,1,0,0) - T(1,0,0,0) = (5,1) - (3,-2) = (2,3) \Rightarrow A_2 = (2,3)\) 3. \(T(e_3) = T((1,1,1,0) - (1,1,0,0)) = T(1,1,1,0) - T(1,1,0,0) = (-1,0) - (5,1) = (-6,-1) \Rightarrow A_3 = (-6,-1)\) 4. \(T(e_4) = T((1,1,1,1) - (1,1,1,0)) = T(1,1,1,1) - T(1,1,1,0) = (2,2) - (-1,0) = (3,2) \Rightarrow A_4 = (3,2)\)

Step 3: Construct the matrix A for T

Now, we will construct the matrix A using the columns found in the previous step: $$A = \begin{pmatrix} 3 & 2 & -6 & 3 \\ -2 & 3 & -1 & 2 \end{pmatrix}$$ So, the matrix for the linear transformation T is: $$T(v) = \begin{pmatrix} 3 & 2 & -6 & 3 \\ -2 & 3 & -1 & 2 \end{pmatrix}v$$

Key Concepts

Linear Transformation Properties

Understanding linear transformations is pivotal in various mathematical and applied disciplines. In essence, linear transformations are functions between vector spaces that preserve vector addition and scalar multiplication. This means that for any vectors \(u, v \in V\) and any scalar \(c\), a linear transformation \(T\) satisfies two crucial properties:

• \(T(u+v) = T(u) + T(v)\)
• \(T(cv) = cT(v)\)

These properties ensure that the structure of the vector space is maintained under the transformation. For example, if we have a transformation \(T: \mathbb{R}^{4} \rightarrow \mathbb{R}^{2}\) like in our exercise, these properties allow us to deduce the outcome of \(T\) on any vector in \(\mathbb{R}^{4}\) by simply knowing \(T\)'s effect on a set of basis vectors.

Matrix Representation of Transformations

To shift from abstract transformations to concrete computations, we anchor the action of linear transformations through matrices. A matrix representation allows us to operationalize linear transformations in terms of array-based computations. For a linear transformation \(T\), we say that a matrix \(A\) represents \(T\) if for every vector \(v\), \(T(v) = Av\).In this framework, each column of the matrix \(A\) corresponds to the image of a basis vector under \(T\). By building such a matrix for a transformation, as shown in our exercise solution, we turn the abstract notion of transforming vectors into the concrete operation of multiplying a matrix by a vector.

Basis Vectors

Basis vectors are the building blocks of vector spaces. In \(\mathbb{R}^n\), standard basis vectors have a '1' in one component and '0' in all others. For example, in \(\mathbb{R}^{4}\), we have:

• \(e_1 = (1,0,0,0)\)
• \(e_2 = (0,1,0,0)\)
• \(e_3 = (0,0,1,0)\)
• \(e_4 = (0,0,0,1)\)

Knowing the effect of a linear transformation \(T\) on each of these basis vectors ultimately determines \(T\)'s effect on the entire space. In other words, by knowing \(T(e_i)\) for all \(i\), we can figure out \(T(v)\) for any vector \(v\) in \(\mathbb{R}^{4}\).

Construction of Transformation Matrix

The construction of a transformation matrix is like piecing together a puzzle. Every column of the matrix reflects how the transformation acts on one of the basis vectors. In our exercise, we deduce the columns of transformation matrix \(A\) using linear properties and given information:

1. Since \(T(e_1) = (3, -2)\), the first column of \(A\) is \(A_1 = (3, -2)\).
2. Subsequent columns are found by applying \(T\) to the difference of given vectors and leveraging linearity:

Deducing Second Column

We have \(T(e_2) = T((1,1,0,0) - (1,0,0,0)) = T(1,1,0,0) - T(1,0,0,0) = (5,1) - (3,-2) = (2,3)\), hence the second column \(A_2 = (2,3)\).

Third Column Determination

Similarly, for \(e_3\) we get \(A_3 = (-6,-1)\) following \(T(e_3)\)'s deduction.

Fourth Column Discovery

Finally, \(A_4 = (3,2)\) comes from \(T(e_4)\).By assembling these columns, we construct the matrix \(A\) that captures the entire action of \(T\) on \(\mathbb{R}^{4}\). This matrix is a compact representation of the linear transformation, enabling us to pinpoint the outcome of \(T(v)\) for any \(v \in \mathbb{R}^{4}\) through matrix multiplication.